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This might be a trivial question (I assume it is at least) but I was wondering if someone could help me out with it. I have an equation given as follows: $${\partial_{t}v + \frac{\sigma^{2}}{2}\frac{\partial_{yy}v}{(\partial_{y}v)^{2}} = \frac{\sigma^{2}}{2}} \tag 1$$

Now, I am told that by taking a variable $w$ to be the inverse of $v$ we now obtain the equation:

$$\partial_{t}w + \frac{\sigma^2}{2}\partial_{xx}w + \frac{\sigma^{2}}{2}\partial_{x}w = 0 \tag 2$$

with

$$\partial_{x}w = \frac{1}{\partial_{y}v} = \frac{1}{s}$$

My question is, how do we obtain the equation $\partial_{t}w + \frac{\sigma^2}{2}\partial_{xx}w + \frac{\sigma^{2}}{2}\partial_{x}w = 0$ ? I assume that for the variable $w$ we must take $w = \frac{1}{v}$ to have the inverse for $v$, but beyond this I have no idea what sort of arrangement of equation $(1)$ will yield equation $(2)$. I think it should be fairly trivial - the researcher I'm working with mentioned that it was obvious/straightforward so I'm sure I'm missing something stupid (Unfortunately I can't talk to them since they're in another country now...). If anyone can help me out with this I'd really appreciate it, thanks in advance.

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  • $\begingroup$ In equation (1), only the variables $y$ and $t$ appear. Is $v$ also function of $x$ that is $v(x,y,t)$ instead of v(y,t) ? . In equation (2), only the variables $x$ and $t$ appear. Is $w$ also function of $y$ that is $w(x,y,t)$ instead of w(x,t) ? . $\endgroup$ – JJacquelin Aug 21 '16 at 9:56
  • $\begingroup$ From the information I'm given $v$ is $v(t, y(t,s))$, while for $w$ no further information is given which also adds to my confusion - it appears after the 'inverse function' $w$ is taken, the variables are swapped or changed... $\endgroup$ – ThePlowKing Aug 21 '16 at 11:25
  • $\begingroup$ Probably there is something missing. May be implicitly given in the context. Without the knowledge of the whole, I doubt that someone could guess. $\endgroup$ – JJacquelin Aug 21 '16 at 16:12

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